Results 1 to 3 of 3

Math Help - Prove that limit doesn't exist

  1. #1
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419

    Prove that limit doesn't exist

    Sorry for all the questions but if there is a recurrence relation such that x_{1} =1, and for n \ge 1, x_{n+1} = 3x_{n}^{2}, how do I show that \lim_{n\to \infty}\,(x_{n}) does not exist. It is obvious, but I do not know how to show it. I'm assuming there is a general formula for this recurrence relation but I cannot find it. Or is there another way to prove it altogether?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Pinkk View Post
    Sorry for all the questions but if there is a recurrence relation such that x_{1} =1, and for n \ge 1, x_{n+1} = 3x_{n}^{2}, how do I show that \lim_{n\to \infty}\,(x_{n}) does not exist. It is obvious, but I do not know how to show it. I'm assuming there is a general formula for this recurrence relation but I cannot find it. Or is there another way to prove it altogether?
    You could just solve the recurrence relation: x_{n+1}=3x_n^2=3\left(3x_{n-1}^2\right)^2=\cdots=3^{2^{n-1}-1}x_1=3^{2^{n-1}-1}. Or, let \ell_n=x_n-n. Clearly we have that \ell_1=1-1\geqslant 0. Now, suppose that \ell_n\geqslant 0. Then, \ell_{n+1}=x_{n+1}-(n+1)=3x_n^2-(n+1) \geqslant 2x_n^2-2n\geqslant 2\left(x_n^2-n^2\right)=2\left(x_n-n\right)\cdot\left(x_n+n\right)\geqslant 3\cdot0\cdot 1=0. It follows by induction that \ell_n\geqslant 0\implies x_n\geqslant n.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    Thanks. It makes sense now but I am no good at solving recurrence relations. But I do not believe I needed to do so anyway. I just figured out that by the recurrence relation, if the limit did exist, it'd have to be either 0 or 1/3. But since every term of that sequence is greater than or equal to 1, then the limit must also be greater than or equal to 1, so obviously it does not exist.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How do you prove that a limit does not exist?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 4th 2009, 06:58 PM
  2. How to Know if Limit Doesn't Exist
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 6th 2009, 01:35 AM
  3. proving limit doesn't exist
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 26th 2009, 10:23 PM
  4. Please help for when a limit doesn't exist
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 20th 2008, 02:46 PM
  5. limit doesn't exist
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 15th 2006, 01:28 PM

Search Tags


/mathhelpforum @mathhelpforum